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Missy science forum beginner
Joined: 02 Jul 2006
Posts: 5

Posted: Mon Jul 10, 2006 11:42 pm Post subject:
PLEASE explain this problem



Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!) 

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David Moran science forum Guru Wannabe
Joined: 13 May 2005
Posts: 252

Posted: Tue Jul 11, 2006 12:04 am Post subject:
Re: PLEASE explain this problem



"Missy" <tawanda.green@yahoo.com> wrote in message
news:1152574969.832779.21250@35g2000cwc.googlegroups.com...
Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
Let n be some positive integer. Then n!=n(n1)(n2)(n3)...(2)(1). So
4!=4(3)(2)(1)=24. Does this help?
Dave 

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Stratocaster science forum addict
Joined: 13 Nov 2005
Posts: 63

Posted: Tue Jul 11, 2006 12:09 am Post subject:
Re: PLEASE explain this problem



"Missy" <tawanda.green@yahoo.com> wrote in message
news:1152574969.832779.21250@35g2000cwc.googlegroups.com...
Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
{Yea! Finally something I can answer lol}
Ok, the Exclamation Mark represents a factorial. A factorial is the product
of all consecutive integers(whole numbers), starting from the integer
signified and goes down the line until 1.
For Example:
5! = 5*4*3*2*1 = 120
3! = 3*2*1 = 6
I'm sure you know this but just in case (I am using * above to indicate
multiplication) ; )
Just for the record
0! = 1
That is the way it is defined. Why would be for another thread.
Hopefully that is all you need to know. 

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jasen science forum beginner
Joined: 28 Jun 2006
Posts: 16

Posted: Tue Jul 11, 2006 12:13 pm Post subject:
Re: PLEASE explain this problem



On 20060710, Missy <tawanda.green@yahoo.com> wrote:
Quote:  Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)

4! = 4 x 3 x 2 x 1
2! = 2 x 1
the answer is 6

Bye.
Jasen 

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matt271829news@yahoo.co. science forum Guru
Joined: 11 Sep 2005
Posts: 846

Posted: Tue Jul 11, 2006 1:06 pm Post subject:
Re: PLEASE explain this problem



Missy wrote:
Quote:  Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)

This is not a "problem", it's an expression. Perhaps you meant to ask
what the notation means, in which case that's been answered. Otherwise,
what is the problem? 

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Darrell science forum addict
Joined: 04 Jun 2005
Posts: 78

Posted: Wed Jul 12, 2006 5:37 am Post subject:
Re: PLEASE explain this problem



"jasen" <jasen@free.net.nz> wrote in message
news:7bb9.44b39606.da153@gonzo.homenet...
Quote:  On 20060710, Missy <tawanda.green@yahoo.com> wrote:
Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
4! = 4 x 3 x 2 x 1
2! = 2 x 1
the answer is 6

I'll just add that sometimes (not necessarily this time, but there will be
times) it is useful to manipulate factorials such that they will cancel (vs.
expanding them out.)
Example: We could rewrite 4! as (4)(3)(2!) and we could further rewrite 4
as (2!)^2 giving
(3)(2!)^3 / (2!)^2
= (3)(2!)
= 6

Darrell 

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Missy science forum beginner
Joined: 02 Jul 2006
Posts: 5

Posted: Wed Jul 12, 2006 10:49 pm Post subject:
thanks



YES it helps.
David Moran wrote:
Quote:  "Missy" <tawanda.green@yahoo.com> wrote in message
news:1152574969.832779.21250@35g2000cwc.googlegroups.com...
Like i have mentioned before I have not had math for some time now and
would like some assistance every now and again. would someone please
explain this problem to me.
4!/(2! × 2!)
Let n be some positive integer. Then n!=n(n1)(n2)(n3)...(2)(1). So
4!=4(3)(2)(1)=24. Does this help?
Dave 


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